Question

Evaluate ∫ 1/√25+9x²
$\huge \pink{∫ 1/√25+9x²}$

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Answer 1

EXPLANATION.

$\sf \implies \displaystyle \int \dfrac{dx}{\sqrt{25 + 9x^{2}} }$

As we know that,

We can write equation as,

Taking 1/3 as common in equation, we get.

$\sf \implies \displaystyle \dfrac{1}{3} \int \dfrac{dx}{\sqrt{\bigg(\dfrac{5}{3}\bigg)^{2} + x^{2} } }$

As we know that,

Formula of :

$\sf \implies \displaystyle \int \dfrac{dx}{\sqrt{x^{2} + a^{2} } } = sinh^{-1} \bigg(\dfrac{x}{a} \bigg) + C \ = log \bigg( x + \sqrt{x^{2} + a^{2} } \bigg) + C$

Using this formula in equation, we get.

$\sf \implies \displaystyle \dfrac{1}{3} \bigg[ log \bigg(x + \sqrt{x^{2} + \bigg(\dfrac{5}{3} \bigg)^{2} } \bigg) \bigg] + C$

$\sf \implies \displaystyle \int \dfrac{dx}{\sqrt{25 + 9x^{2}} } \ = \dfrac{1}{3} \bigg[log \bigg(x + \sqrt{x^{2} + \bigg(\dfrac{5}{3} \bigg)^{2} } \bigg) \bigg] + C$

                                                                                                                       

MORE INFORMATION.

Integration of trigonometric functions.

(1) = ∫dx/a + bsin²x.

(2) = ∫dx/a + bcos²x.

(3) = ∫dx/acos²x + b sinx.cosx + csin²x.

(4) = ∫dx/(a sin x + b sin x)².

Divide numerator and denominator by cos²x in all such type of integrals and then put tan x = t.

Answer 2

We have,

$\sf \: I = ∫ \bf\frac{1}{ \sqrt{25 + 9x^{2} } \: }$

$\sf \: I = \frac{1}{3} ∫ \frac{1}{ \bf\sqrt{ (\frac{5}{3} )^{2} + {x}^{2} } }$

So, applying formula,

$\sf∫ \frac{1}{ \sqrt{ {a}^{2} + {x}^{2} } } = In \:[ x + \sqrt{ {x}^{2} + {a}^{2} \: ] }$

Therefore,

$\sf \: I= \frac{1}{3} \: In \: [ \: x + \sqrt{ {x}^{2} + ( \frac{5}{3})^{2} \: \: ] } + C$

Hence, this is the answer ツ